heureka

$$\\ \left[
5-{3\times12\over(5\times5+15)}
+(3\times0+40)
\right]
\times(5\times5
\times{3\over15}
)\\\\
=(5-{36\over40}+40})\times(5\times {15\over15})\\\\
=(45-{9\over10}})\times5\\\\
=45\times5-{45\over10}}\\\\
=225-4.5\\\\
=220.5$$

how to convert 486mm to m:

A parabola is defined by the equation . In which direction will the parabola open?

$$\boxed{y=ax^2+bx+c} \qquad \mbox{ parabola}$$

a > 0     opens upward 

a < 0     opens downward

$$\\x^2={3 \over 4}y\\\\
y=\underbrace{{4 \over 3}}_{ a > 0} x^2$$

a > 0 parabola opens upward

e^2x + 1 = 900   ?

$$\begin{array}{rcrc}
e^{2x+1} &=& 900 & \quad | \quad \ln{}\\
\ln{(e^{2x+1} )}&=&\ln{( 900 )} & \\
(2x+1)*\ln{( e )}&=&\ln{( 900 )} & \quad |\quad \ln{(e)} = 1 \quad !\\
2x+1 &=&\ln{( 900 )} & \\
2x&=&\ln{( 900 )} - 1& \\
x&=&{ \ln{(900)} -1 \over 2 } &\\
x&=&2.90119738166&\end{array}$$

9^−x/200 = 2  ?

$$\begin{array}{rcrc}
9^{
(-{x\over 200}) }
&=& 2 & \quad | \quad \ln{}\\

\ln{(
9^{
(-{x\over 200}) }
)
}&=&\ln{(2 )} & \\
(-{x\over 200}) }
*\ln{(9 )}&=&\ln{(2 )} & | \quad *\quad -({200\over \ln{(9)}})\\
x&=&-200*{\ln{(2)}\over\ln{(9)}} &\\
x&=&-63.0929753571&
\end{array}$$

7^0.9x = 6   ?

$$\begin{array}{rcrc}
7^{0.9x} &=& 6 & \quad | \quad \ln{}\\
\ln{(
7^{0.9x} )}
&=&
\ln{(
6 )} & \\
0.9*x*\ln{(
7 )}
&=&
\ln{(
6 )} & \\
x&=&{
\ln{(6)}
\over
0.9 *\ln{(7)}
}
&\\
x&=&1.02309135685&
\end{array}$$  

x ² 3 ^x −  4( 3 ^x) = 0    ?

$$\\x^2 3^x-4(3^x)=0 \\
3^x
(
x^2-4
)
=0 \\
3^x
(
x-2
)
(
x+2
)
=0\\\\
\underbrace{3^x}_{=0}
\times(
\underbrace{x-2}_{=0}
)
\times(
\underbrace{x+2}_{=0}
)
=0 \\\\
\boxed{3^x=0} \quad | \quad \ln{} \\\\
\ln{(3^x)} = \ln{(0)} \\
x\ln{(3)} = \ln{(0)} \\
x=
{
\ln{(0)}
\over \ln{(3)} }
} \quad | \quad \ln{(0)} \mbox{ no solution !}\\\\
\boxed{x-2=0} \quad \Rightarrow \quad \boxed{x=x_1=2}\\\\
\boxed{x+2=0} \quad \Rightarrow \quad \boxed{x=x_2=-2}$$

e^6x + 5e^3x − 14 = 0 ?

$$\\e^{6x}=e^{3x}e^{3x} \quad | \quad \mbox{ set }\quad \boxed{z=e^{3x}}\\
z^2 +5z -14=0\\
\underbrace{1}_{a=1}z^2 \underbrace{+5}_{b=5}z \underbrace{-14}_{c=-14} =0\\
\boxed{z_{1,2}=
{
-b\pm\sqrt{b^2-4ac}
\over
2a
}
}\\\\
z_{1,2}=
{ -5\pm\sqrt{25-4*1*(-14)}
\over
2*1 } \\\\
z_{1,2}=
{ -5\pm\sqrt{25+56}
\over
2} \\\\
z_{1,2}=
{ -5\pm9\over
2} \\\\
\boxed{z_1= 2 \quad z_2 = -7} \\\\
e^{3x}=z \quad | \quad \ln\\\\
\ln{(e^{3x})}=\ln{(z)}\\\\
3x\ln{(e)}=\ln{(z)} \quad | \quad \ln{(e)}= 1 \quad !\\\\
\boxed{x={1\over 3}\ln{(z)}} \\\\$$

$$\\x=x_1={
\ln{(2)}\over 3}
}
=0.23104906019\\ x=x_2={
\ln{(-7)}\over 3}
}
= \mbox{no solution !}\\\\
\boxed{x=0.23104906019}$$

e^2x − 9e^x + 8 = 0  ?

 $$\\e^{2x}=e^xe^x \quad | \quad \mbox{ set }\quad \boxed{z=e^x}\\
z^2 -9z +8=0\\
\underbrace{1}_{a=1}z^2 \underbrace{-9}_{b=-9}z \underbrace{+8}_{c=8} =0\\
\boxed{z_{1,2}=
{
-b\pm\sqrt{b^2-4ac}
\over
2a
}
}\\\\
z_{1,2}=
{ 9\pm\sqrt{81-4*1*8}
\over
2*1 } \\\\
z_{1,2}=
{ 9\pm\sqrt{81-32}
\over
2} \\\\
z_{1,2}=
{ 9\pm7\over
2} \\\\
\boxed{z_1= 8 \quad z_2 = 1} \\\\
e^x=z \quad | \quad \ln\\\\
\ln{(e^x)}=\ln{(z)}\\\\
x\ln{(e)}=\ln{(z)} \quad | \quad \ln{(e)}= 1 \quad !\\\\
\boxed{x=\ln{(z)}} \\\\$$

$$\\x=x_1=\ln{(8)}=2.07944154168\\
x=x_2=\ln{(1)}=0\\
\boxed{x_1=2.07944154168 \qquad x_2=0}$$

why is 1 not a prime number? why?

Prime numbers are never divisible by other prime numbers. If was 1 a prime number, would be valid this no more, because every number is divisible by 1